anonymous
  • anonymous
If A=B, then AnB=AuB
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
DebbieG
  • DebbieG
ok... and? :)
DebbieG
  • DebbieG
Welcome to Open Study, by the way! :) What's the question? are you supposed to prove this?
anonymous
  • anonymous
That it does always, never, or sometimes?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

DebbieG
  • DebbieG
Ah, ok... well, what do you think?
DebbieG
  • DebbieG
It's kind of hard to Venn diagram it, since A=B are just the same circle, right? so if A=B then everything that's in A is in B, and everything that's in B is in A (there is no difference between the 2, no elements in 1 that aren't in the other). so what does that mean for the union? What does that mean for the intersection?
anonymous
  • anonymous
Not sure. Isn't that if A=B then AnB=B and AuB=B, so is it never?
DebbieG
  • DebbieG
welll..... But the question is: Does A=B imply that AnB=AuB? You just said that A=B implies that AnB=B and AuB=B... which I'll agree with..... So, DOES it mean that AnB=AuB?
DebbieG
  • DebbieG
By the way, A=B also implies that AnB=A and AuB=A. Remember, they are the SAME sets.
anonymous
  • anonymous
So it is always?
DebbieG
  • DebbieG
Think about what it MEANS, that A=B.... don't just try to "fit" this into some rule you've already learned. A=B means they have EXACTLY the same elements. So what's in A{intersect}B ?? Everything that is BOTH sets, right? What in A{union}B ?? Everything that is in EITHER set, right? Well, THEY ARE THE SAME SETS. So, isn't everything that's in BOTH, also in EITHER?
DebbieG
  • DebbieG
Yes, I would think always. Because they are the same sets, so every element that is in EITHER set, it in BOTH sets. And every element in BOTH sets is in EITHER set. They are the same. :)
anonymous
  • anonymous
I know intersections are the same numbers that they have in common and and union is of all the numbers combined but not repeated.
DebbieG
  • DebbieG
Try a few examples to convince yourself... that doesn't PROVE it, but it will help you gain intuition. A={1,3,5,7,8}, .B={1,3,5,7,8} What's the union? what's the intersection? See what I mean? :)
anonymous
  • anonymous
The union is {1,3,5,7,8} and so if the intersection. Right?
anonymous
  • anonymous
*is
DebbieG
  • DebbieG
Right... |dw:1378686614862:dw| Now, imagine that picture if A=B... the circles "slide" over so that they are exactly on top of each other (the sets are the same). Think about what happens to the intersection area, and the union area. BOTH just become identical to the circle.
DebbieG
  • DebbieG
And yes, that's right. Like I said, doesn't PROVE that it's "always" but helps you get some intuition around it.
anonymous
  • anonymous
Can you help me with evaluate the expression 5^-1/5^0?
DebbieG
  • DebbieG
I will, but in the future please post each question separately, not in the same thread. It just works better. :) So is the expression \(\Large \dfrac{5^{-1}}{5^0}\) ?
DebbieG
  • DebbieG
You just need a couple of rules for exponents. Actually, there are a couple of ways to go about this one... but you will need: \(\Large a^{-1}=\dfrac{1}{a}\)
anonymous
  • anonymous
Ok. Yes.
DebbieG
  • DebbieG
And also \(a^0=1\) for any base, a. Does that sound familiar?
DebbieG
  • DebbieG
So apply those two rules to your expression: \(\Large \dfrac{5^{-1}}{5^0}\) OR, alternatively, you COULD just use this rule: \(\Large \dfrac{a^{m}}{a^n}=a^{m-n}\)
anonymous
  • anonymous
Yes.
DebbieG
  • DebbieG
OK, you should have everything you need there to simplify the expression. Go ahead and try.
anonymous
  • anonymous
I can't remember how to begin. I'm trying to help my daughter who is on homebound due to surgery.
anonymous
  • anonymous
It has been years since I have had Algebra.
anonymous
  • anonymous
I have that it is 1/5 / 1
DebbieG
  • DebbieG
that's a bit hard to interpret... lol.. but I think you have it, if you mean: \[\Large \dfrac{ \dfrac{ 1 }{ 5 } }{ 1 }\] but notice that you can "simplify" it, since there is no need to write anything over the den'r of 1. That is, a/1 = 1. so you just write: \(\Large \dfrac{ 1 }{ 5 } \) :)
DebbieG
  • DebbieG
Also, you get that simply if you use the rule I gave above: \(\Large \dfrac{a^{m}}{a^n}=a^{m-n}\) Then you have: \(\Large \dfrac{5^{-1}}{5^0}=5^{-1-0}=5^{-1}= \dfrac{1}{5}\)
anonymous
  • anonymous
Thank you so much :) It is hard to teach a child Algebra when you have been out of school for so long and have forgotten everything yourself and are having to relearn yourself.

Looking for something else?

Not the answer you are looking for? Search for more explanations.